# Electronics - Transformers - Discussion

Discussion Forum : Transformers - General Questions (Q.No. 10)
10.

What is the power dissipated in the primary of the transformer in the given circuit?

25 mW
500 mW
12.5 W
62.5 W
Explanation:
No answer description is available. Let's discuss.
Discussion:
9 comments Page 1 of 1.

Sohail khan said:   5 years ago
Power will be 62.5 watt, not mw and current will also in ampere not in mili because resistance is in kilo ohm not just 1 ohm.

Akshay said:   7 years ago
How, v2 = 250 please explain.

PrivUsr said:   8 years ago
A simple explanation - primary power equal to secondary power in a transformer. Assuming this holds, then.

V2 = 50 * 5 = 250 V.
P2 = V**2 / R = 62.5 W.

For a transformer, the AC voltage & current could be different but power remain equal.

Ankur said:   10 years ago
v2/v1=n2/n1. So v2=250v,
According to ohm's law v = ir.

i2 = v2/r2 = 250mA.
i1/i2 = v2/v1. So i1=1250 mA.

p=vi So pin=v1*i1=50*1250 = 62.5W.

Input power and Output power should be the same, therefore calculate secondary power by v^2/r.

250^2/1000 = 62.5w.

We have,

Ip/Is = Ns/Np =>Ip = (Ns/Np)*Is ------(i).

Is = Vs/Rl--------(ii).

where Vs = (Ns/Np)*Vp.

Which gives Is = 250mA.

Thus we ve from eqn (i).

Ip = 1.25A.

Therefore, Pp = V*Ip = 62.5W.

Please check this way is correct or not sir/mam r= 1kohm 1*1000, v=250.

p=v^/r
=250*250/1000
=62.5 w

It is very simple
r1/r2=sqr(n1/n2) so r1=?
r2=1k
r1=r2*sqr(n1/n2)
r1=40ohm
p=v^2/r
p=62.5W